<u>Given:</u>
Mass of H2O2 solution = 5.02 g
Mass of H2O2 = 0.153 g
<u>To determine: </u>
The % H2O2 in solution
<u>Explanation:</u>
Chemical reaction-
2H2O2(l) → 2H2O(l) + O2(g)
Mass % of a substance in a solution = (Mass of the substance/Mass of solution) * 100
In this case
% H2O2 = (Mass H2O2/Mass of solution)* 100 = (0.153/5.02)*100 = 3.05%
Ans: % H2O2 in the solution = 3.05%
Answer:
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Answer:
Given the activity series of elements, the reaction that will occur is: 2AgNO3 + Ni -----> Ni(NO3)2 + 2Ag
Answer:
0.42 g
Explanation:
<u>We have: </u>
pH = 12.10 (25 °C)
V = 800.0 mL = 0.800 L
To find the mass of sodium hydroxide (NaOH) we can use the pH:
![pOH = -log ([OH^{-}])](https://tex.z-dn.net/?f=%20pOH%20%3D%20-log%20%28%5BOH%5E%7B-%7D%5D%29%20)
![[OH]^{-} = 10^{-pOH} = 10^{-1.90} = 0.013 M](https://tex.z-dn.net/?f=%5BOH%5D%5E%7B-%7D%20%3D%2010%5E%7B-pOH%7D%20%3D%2010%5E%7B-1.90%7D%20%3D%200.013%20M)
Now, we can find the number of moles (η) of OH:
![\eta = ([OH]^{-})*V = 0.013 mol/L * 0.800 L = 1.04 \cdot 10^{-2} moles](https://tex.z-dn.net/?f=%5Ceta%20%3D%20%28%5BOH%5D%5E%7B-%7D%29%2AV%20%3D%200.013%20mol%2FL%20%2A%200.800%20L%20%3D%201.04%20%5Ccdot%2010%5E%7B-2%7D%20moles)
Since we have 1 mol of OH in 1 mol of NaOH, the number of moles of NaOH is equal to 1.04x10⁻² moles.
Finally, with the number of moles we can find the mass of NaOH:
<em>Where M is the molar mass of NaOH = 39.9 g/mol </em>
Therefore, the mass of sodium hydroxide that the chemist must weigh out in the second step is 0.42 g.
I hope it helps you!
Answer : To determine enthalpy of solution for a solid with δH values.
One can use the equation as H (reaction) = H (products) - H (reactants)
If δH values are known one can simply substitute them in the above equation and get the enthalpy of the required solution.
